[Soln] 2014 Singapore MO (Senior Rd 1) Problem 16

Posted on 16 Sep 2014 by kjytay

2014 SMO (Senior-Rd1) 16. Evaluate the sum

\displaystyle\frac{3! + 4!}{2(1! + 2!)} + \frac{4! + 5!}{3(2! + 3!)} + \dots + \frac{12! + 13!}{11(10! + 11!)}.

Recall the definition of the factorial function: for non-negative integer n,

n! = \begin{cases} 1 &\text{ for } n = 0, \\ n \cdot (n-1) \cdot \dots \cdot 1 &\text{ for } n \geq 1. \end{cases}

When faced with a series of many terms, try to come up with an expression for the general \boldsymbol{n^{th}} term. In this question, the required sum is the sum of

\displaystyle\frac{(n+2)! + (n+3)!}{(n+1)[n! +(n+1)!]}

for n = 1, 2, \dots, 10. This looks like a term that can be simplified. Especially with fractions, a possible strategy is to pull out as many factors as possible in the hope that there can be some common factors in the numerator and denominator. These can be cancelled out. By pulling out common factors in the numerator:

\begin{aligned} (n+2)! + (n+3)! &= (n+2)! \cdot [1 + (n + 3)] \\ &= (n+2)! \cdot (n+4) \end{aligned}

and in the denominator:

\begin{aligned} (n+1)[n! + (n+1)!] &= (n+1) \cdot n! \cdot [1 + (n+1)] \\ &= (n+1) \cdot n! \cdot (n+2) \\ &= (n+2)! \end{aligned}

The original expression can be greatly simplified:

\displaystyle\frac{(n+2)! + (n+3)!}{(n+1)[n! +(n+1)!]} = \frac{(n+2)! \cdot (n+4)}{(n+2)!} = n + 4.

Hence, the required sum is

\begin{aligned} (1+4) + (2+4) + \dots + (10 + 4) &= 5 + 6 + \dots + 14 \\ &= (1 + 2 + \dots + 14) - (1 + 2 + 3 + 4) \\ &= \frac{14 \cdot 15}{2} - 10 \\ &= 105 - 10 \\ &= \boldsymbol{95}. \end{aligned}

The answer is 95.

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